m4l30.pdf Plane Frames

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Module 
4 
 
Analysis of Statically 
Indeterminate 
Structures by the Direct 
Stiffness Method 
Version 2 CE IIT, Kharagpur 
 
 
 
 
 
 
 
 
 
Lesson 
30 
The Direct Stiffness 
Method: Plane Frames 
Version 2 CE IIT, Kharagpur 
 
Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive plane frame member stiffness matrix in local co-ordinate system. 
2. Transform plane frame member stiffness matrix from local to global co-
ordinate system. 
3. Assemble member stiffness matrices to obtain the global stiffness matrix of 
the plane frame. 
4. Write the global load-displacement relation for the plane frame. 
5. Impose boundary conditions on the load-displacement relation. 
6. Analyse plane frames by the direct stiffness matrix method. 
 
 
 30.1 Introduction 
 In the case of plane frame, all the members lie in the same plane and are 
interconnected by rigid joints. The internal stress resultants at a cross-section of 
a plane frame member consist of bending moment, shear force and an axial 
force. The significant deformations in the plane frame are only flexural and axial. 
In this lesson, the analysis of plane frame by direct stiffness matrix method is 
discussed. Initially, the stiffness matrix of the plane frame member is derived in 
its local co-ordinate axes and then it is transformed to global co-ordinate system. 
In the case of plane frames, members are oriented in different directions and 
hence before forming the global stiffness matrix it is necessary to refer all the 
member stiffness matrices to the same set of axes. This is achieved by 
transformation of forces and displacements to global co-ordinate system. 
 
 
30.2 Member Stiffness Matrix 
Consider a member of a plane frame as shown in Fig. 30.1a in the member co-
ordinate system ' . The global orthogonal set of axes '' zyx xyz is also shown in the 
figure. The frame lies in the xy plane. The member is assumed to have uniform 
flexural rigidity EI and uniform axial rigidity EA for sake of simplicity. The axial 
deformation of member will be considered in the analysis. The possible 
displacements at each node of the member are: translation in - and - 
direction and rotation about - axis. 
'x 'y
'z
 
Version 2 CE IIT, Kharagpur 
 
 
 
Thus the frame members have six (6) degrees of freedom and are shown in 
Fig.30.1a. The forces acting on the member at end j and are shown in Fig. 
30.1b. The relation between axial displacement and axial forces is derived in 
chapter 24. Similarly the relation between shear force, bending moment with 
translation along axis and rotation about axis are given in lesson 27. 
Combining them, we could write the load-displacement relation in the local co-
ordinate axes for the plane frame as shown in Fig 30.1a, b as, 
k
'y 'z
 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
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460260
61206120
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'
u
u
u
u
u
u
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
AE
L
AE
q
q
q
q
q
q
zzzz
zzzz
zzzz
zzzz
 (30.1a) 
 
This may be succinctly written as 
 { } [ ]{ }''' ukq = (30.1b) 
 
 
where is the member stiffness matrix. The member stiffness matrix can also 
be generated by giving unit displacement along each possible displacement 
degree of freedom one at a time and calculating resulting restraint actions. 
[ ]'k
 
 
30.3 Transformation from local to global co-ordinate system 
30.3.1 Displacement transformation matrix 
In plane frame the members are oriented in different directions and hence it is 
necessary to transform stiffness matrix of individual members from local to global 
co-ordinate system before formulating the global stiffness matrix by assembly. In 
Fig. 30.2a the plane frame member is shown in local coordinate axes zyx ′′′ and 
in Fig. 30.2b, the plane frame is shown in global coordinate axes xyz . Two ends 
of the plane frame member are identified by j and . Let and 
 be respectively displacements of ends 
k 321 ',',' uuu
654 ',',' uuu j and of the member in local 
coordinate system ' . Similarly and respectively are 
displacements of ends 
k
'' zyx 321 ,, uuu 654 ,, uuu
j and k of the member in global co-ordinate system. 
Version 2 CE IIT, Kharagpur 
 
 
 
Let θ be the angle by which the member is inclined to global x -axis. From 
Fig.30.2a and b, one could relate to as, 321 ',',' uuu 321 ,, uuu
 
θθ sincos' 211 uuu += (30.2a) 
 
θθ cossin' 212 uuu +−= (30.2b) 
 
33' uu = (30.2c) 
 
This may be written as, 
 
 
Version 2 CE IIT, Kharagpur 
 
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6
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100000
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'
'
'
'
'
u
u
u
u
u
u
lm
ml
lm
ml
u
u
u
u
u
u
 (30.3a) 
 
Where, θcos=l and θsin=m . 
 
This may be written in compact form as, 
 { } [ ]{ }uTu =' (30.3b) 
 
 
In the above equation, [ is defined as the displacement transformation matrix 
and it transforms the six global displacement components to six displacement 
components in local co-ordinate axes. Again, if the coordinate of node 
]T
j is 
 and coordinate of node are ( 11, yx ) k ( )22 , yx , then, 
 
L
xxl 12cos −== θ and 
L
yym 12sin −== θ . 
 
Where ( ) ( )212212 yyxxL −+−= (30.4) 
 
 
 
 
 
 
 
 
 
 
 
 
Version 2 CE IIT, Kharagpur 
 
30.3.2 Force displacement matrix 
 
 
 
Let and be respectively the forces in member at nodes 321 ',',' qqq 654 ',',' qqq j 
and as shown in Fig. 30.3a in local coordinate system. and 
are the forces in members at node 
k 321 ,, ppp
654 ,, ppp j and respectively as shown in 
Fig. 30.3b in the global coordinate system. Now from Fig 30.3a and b, 
k
 
θθ sin'cos' 211 qqp −= (30.5a) 
 
θθ cos'sin' 212 qqp += (30.5b) 
 
Version 2 CE IIT, Kharagpur 
 
33 'qp = (30.5c) 
 
Thus the forces in global coordinate system can be related to forces in local 
coordinate system by 
 
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−
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6
5
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q
q
q
qq
q
lm
ml
lm
ml
p
p
p
p
p
p
 (30.6a) 
 
Where, θcos=l and θsin=m . 
 
This may be compactly written as, 
 
{ } [ ] { }'qTp T= (30.6b) 
 
30.3.3 Member global stiffness matrix 
From equation (30.1b), we have 
 { } [ ]{ }''' ukq = 
 
Substituting the above value of { }'q in equation (30.6b) results in, 
 
{ } [ ] [ ]{ }'' ukTp T= (30.7) 
 
Making use of equation (30.3b), the above equation may be written as 
 
 
{ } [ ] [ ][ ]{ }uTkTp T '= (30.8) 
or 
 { } [ ]{ }ukp = (30.9) 
 
The equation (30.9) represents the member load-displacement relation in global 
coordinate system. The global member stiffness matrix [ ]k is given by, 
 
[ ] [ ] [ ][ ]TkTk T '= (30.10) 
Version 2 CE IIT, Kharagpur 
 
 
After transformation, the assembly of member stiffness matrices is carried out in 
a similar procedure as discussed for truss. Finally the global load-displacement 
equation is written as in the case of continuous beam. Few numerical problems 
are solved by direct stiffness method to illustrate the procedure discussed. 
 
Example 30.1 
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method. 
Assume and . The flexural rigidity 441033.1;200 mIGPaE ZZ
−×== 204.0 mA =
EI and axial rigidity EA are the same for both the beams. 
 
 
 
Solution: 
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The 
numbering of joints and members are also shown in Fig. 30.3b. Each node has 
three degrees of freedom. Degrees of freedom at all nodes are also shown in the 
figure. Also the local degrees of freedom of beam element are shown in the 
figure as inset. 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Formulate the element stiffness matrix in local co-ordinate system and then 
transform it to global co-ordinate system. The origin of the global co-ordinate 
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given. 
 
Member 1: °== 90;6 θmL node points 1-2; 0=l and 1=m . 
Version 2 CE IIT, Kharagpur 
 
[ ] [ ] [ ]1 'Tk T k T⎡ ⎤ =⎣ ⎦
3 3 3 3
6 6
3 3 3
1
3 3 3 3
6 6
3 3 3
 1 2 3 4 5 6
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 17.78 10 4.44 10 0 8.88 10
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 8.88 10 4.44 10 0 17.
k
× × × ×
× − ×
× × × ×⎡ ⎤ =⎣ ⎦ × × × ×
− × ×
× × × 3
1
2
3
4
5
678 10
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥×⎢ ⎥⎣ ⎦
3
 (1) 
 
Member 2: °== 0;4 θmL ; node points 2-3 ; 1=l and 0=m . 
 [ ] [ ] [ ][ ]
9
8
7
6
5
4
1066.26101001088.810100
1010105010101050
00100.200100.2
1088.8101001066.2610100
1010105010101050
00100.200100.2
987654
'
3333
3333
66
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66
2
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⎣
⎡
××−××
×−××−×−
××−
××−××
××−××
×−×
=
= TkTk T
 
 (2) 
The assembled global stiffness matrix [ ]K is of the order 99× . Carrying out 
assembly in the usual manner, we get, 
Version 2 CE IIT, Kharagpur 
 
[ ]
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−
−−−
−
−−
−−
−−
−
−
−−−
=
66.2610033.13100000
10501050000
002000002000000
33.1310044.441044.488.8044.4
1050103.1338003.13330
00200044.405.200144.4048.1
00088.8044.477.17044.4
00003.1333003.13330
00044.4048.144.4048.1
K (3) 
 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Version 2 CE IIT, Kharagpur 
 
 
 
The load vector corresponding to unconstrained degrees of freedom is (vide 
30.4d), 
 
{ }
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
−=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=
6
24
12
6
5
4
p
p
p
pk (4) 
 
In the given frame constraint degrees of freedom are . 
Eliminating rows and columns corresponding to constrained degrees of freedom 
from global stiffness matrix and writing load-displacement relationship for only 
unconstrained degree of freedom, 
987321 ,,,,, uuuuuu
Version 2 CE IIT, Kharagpur 
 
 
 
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=
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−
−
6
5
4
3
44.441044.4
103.13380
44.405.2001
10
6
24
12
u
u
u
 (5) 
 
Solving we get, 
 
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⎪⎪⎬
⎫
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⎧
×
×
×
=
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⎫
⎪⎪⎩
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⎧
3-
5-
-6
6
5
4
100.13-
101.695-
106.28
u
u
u
 (6) 
 
6 5
4 56.28 10 m., 1.695 10u u
− −= × = − × 
 
Let be the support reactions along degrees of freedom 
 respectively (vide Fig. 30.4e). Support reactions are calculated by 
987321 ,,,,, RRRRRR
9,8,7,3,2,1
 
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33.13100
1050
002000
88.8044.4
03.1333
44.4048.1
10
654
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
 
 
 
 
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−
−
−+
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=
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⎬
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⎪⎪
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⎪
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⎧
92.25
40.25
57.12
85.16
59.22
42.11
92.1
40.1
57.12
14.1
59.22
57.0
24
24
0
18
0
12
9
8
7
3
2
1
R
R
R
R
R
R
 (7) 
Version 2 CE IIT, Kharagpur 
 
Example 30.2 
Analyse the rigid frame shown in Fig 30.5a by direct stiffness matrix method. 
Assume and . The flexural 
rigidity 
5 4200 GPa ; 1.33 10 mZZE I
−= = × 201.0 mA =
EI and axial rigidity EA are the same for all beams. 
 
 
 
Solution: 
The plane frame is divided in to three beam elements as shown in Fig. 30.5b. 
The numbering of joints and members are also shown in Fig. 30.5b. The possible 
degrees of freedom at nodes are also shown in the figure. The origin of the 
global co- ordinate system is taken at A (node 1). 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Now formulate the element stiffness matrix in local co-ordinate system and then 
transform it to global co-ordinate system. In the present case three degrees of 
freedom are considered at each node. 
 
Member 1: °== 90;4 θmL ; node points 1-2 ; 2 1 0x xl
L
−= = and 
2 1 1y ym
L
−= = . 
 
The following terms are common for all elements. 
 
5 2
2
65 10 kN/m; 9.998 10 kNAE EI
L L
= × = × 
 
2 3
3
12 44.999 10 kN/m; 2.666 10 kN.mEI EI
L L
= × = × 
Version 2 CE IIT, Kharagpur 
 
 
[ ] [ ] [ ][ ]
6
5
4
3
2
1
1066.201011033.10101
0105001050
10101050.010101050.0
1033.101011066.20101
010500105010101050.010101050.0
654321
'
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55
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55
3333
1
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⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
××××−
××−
××××−
××××−
×−×
×−×−×−×
=
= TkTk T
 
 (1) 
 
Member 2: °== 0;4 θmL node points 2-3; 1=l and 0=m . 
 [ ] [ ] [ ][ ]
9
8
7
6
5
4
10666.210101033.11010
101105.00101105.00
00100.500100.5
1033.1101010666.21010
101105.00101105.00
00100.500100.5
987654
'
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3333
66
3333
3333
65
2
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⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
××−××
×−××−×−
××−
××−××
××−××
×−×
=
= TkTk T
 
 (2) 
 
 
Member 3: °== 270;4 θmL ; node points 3-4 ; 2 1 0x xl
L
−= = and 
2 1 1y ym
L
−= = − . 
 
 
Version 2 CE IIT, Kharagpur 
 
 
[ ] [ ] [ ][ ]
12
11
10
9
8
7
1066.201011033.10101
0105001050
10101050.010101050.0
1033.101011066.20101
0105001050
10101050.010101050.0
121110987
'
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55
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55
3333
3
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
××−××
××−
×−××−×−
××−××
×−×
××−××
=
= TkTk T
 (3) 
 
The assembled global stiffness matrix [ ]K is of the order 1212× . Carrying out 
assembly in the usual manner, we get, 
 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢
⎣
⎡
−
−
−−−
−−
−−−−
−−
−−
−−
−−
−
−
−−−
×=
66.20133.100.1000000
0500005000000000
105.00.105.0000000
33.100.133.50.10.133.10.10000
050000.15.50000.15.00000
0.105.00.105.50000500000
00033.10.1033.50.10.133.100.1
0000.150.000.15.500005000
000005000.105.5000.1050.0
00000033.100.166.200.1
0000000500005000
0000000.1050.00.1050.0
103K
 
 (4) 
 
Version 2 CE IIT, Kharagpur 
 
 
 
 
Version 2 CE IIT, Kharagpur 
 
The load vector corresponding to unconstrained degrees of freedom is, 
 
 
{ }
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
24
24
0
24
24
10
9
8
7
6
5
4
p
p
p
p
p
p
pk (5) 
 
In the given frame, constraint (known) degrees of freedom are 
. Eliminating rows and columns corresponding to constrained 
degrees of freedom from global stiffness matrix and writing load displacement 
relationship, 
121110321 ,,,,, uuuuuu
 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
−
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−
9
8
7
6
5
4
3
33.51133.110
15.500015.00
105.50000500
33.10.1033.50.10.1
0.15.000.15.5000
005000.105.500
10
24
24
0
24
24
10
u
u
u
u
u
u
 (6) 
 
Solving we get, 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
×
×
×
×
×
×
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
3-
5-
2-
3-
5-
-2
9
8
7
6
5
4
103.85
105.65-
101.43
108.14-
103.84-
1043.1
u
u
u
u
u
u
 (7) 
 
Version 2 CE IIT, Kharagpur 
 
Let be the support reactions along degrees of freedom 
 respectively. Support reactions are calculated by 
121110321 ,,,,, RRRRRR
12,11,10,3,2,1
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−
−
−−
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
9
8
7
6
5
4
3
12
11
10
3
2
1
12
11
10
3
2
1
33.100.1000
05000000
0.105.0000
00033.100.1
00005000
0000.105.0
10
987654
u
u
u
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
 
 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
42.19
28.28
99.10
43.3
71.19
99.0
42.19
28.28
99.10
43.3
71.19
99.0
0
0
0
0
0
0
12
11
10
3
2
1
R
R
R
R
R
R
 (8) 
 
 
Summary 
In this lesson, the analysis of plane frame by the direct stiffness matrix method is 
discussed. Initially, the stiffness matrix of the plane frame member is derived in 
its local co-ordinate axes and then it is transformed to global co-ordinate system. 
In the case of plane frames, members are oriented in different directions and 
hence before forming the global stiffness matrix it is necessary to refer all the 
member stiffness matrices to the same set of axes. This is achieved by 
transformation of forces and displacements to global co-ordinate system. In the 
end, a few problems are solved to illustrate the methodology. 
Version 2 CE IIT, Kharagpur 
 
	Analysis of Statically Indeterminate Structures by the Direct Stiffness Method
	The Direct Stiffness Method: Plane Frames
	Instructional Objectives
	30.1 Introduction
	30.2 Member Stiffness Matrix
	30.3 Transformation from local to global co-ordinate system
	30.3.1 Displacement transformation matrix
	30.3.2 Force displacement matrix
	30.3.3 Member global stiffness matrix
	Summary